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Thread: limits

  1. #1
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    limits

    Find the Limit.

    lim x->oo sqrt(x^2+1)

    oo=infinity
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  2. #2
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    Quote Originally Posted by johntuan View Post
    Find the Limit.

    lim x->oo sqrt(x^2+1)

    oo=infinity
    $\displaystyle \sqrt{x^2+1} \geq \sqrt{x^2+0} = \sqrt{x^2} = x$ for $\displaystyle x>0$.

    And $\displaystyle \lim_{x\to \infty} x = \infty$ so $\displaystyle \lim_{x\to \infty}\sqrt{x^2+1} = \infty$.
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  3. #3
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    how come the 1 turns to a 0?
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  4. #4
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    Quote Originally Posted by johntuan View Post
    how come the 1 turns to a 0?
    It does not. It use "$\displaystyle \geq$" there.

    Basically I am showing that $\displaystyle \sqrt{x^2+1}$ is larger than $\displaystyle x$. And since $\displaystyle x$ goes to $\displaystyle \infty$ must mean $\displaystyle \sqrt{x^2+1}$ goes to $\displaystyle \infty$.
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